An ideal gas expands isothermally at 510 K from 8 L to 24 L with 0.2 moles . What is the work done by the gas? ( R = 8.3
**Isochoric process** constant volume ΔV=0, work W=0, first law ΔU = Q, all heat goes to internal energy, P/T = constant from ideal gas law P V = n R T at constant V, pressure proportional to temperature, P₁/T₁ = P₂/T₂, e.g., heating gas in rigid container pressure rises proportionally to T. For isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.2 , R = 8.3 , T = 510 , V₂ = 24 , V₁ = 8 . W = 0.2 × 8.3 × 510 × ln((24)/(8)) = 846.6 × ln(3) . ln(3) ≈ 1.0986 , W ≈ 846.6 × 1.0986 ≈
Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature